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Question Number 54808 by turbo msup by abdo last updated on 11/Feb/19

l e t p (x) = (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})

w i t h n i n t e g r n a t u r a l

1) f i n d a s i m p l e f o r m o f p (x)

2) f i n d r o o t s o f p (x) a n d d e c o m p o s e

p (x) i n s i d e C [x]

3) c a l c u l a t e \int_{0}^{1} p (x) d x

4) d e c o m p o s e t h e f r a c t i o n

F (x) = \frac{1}{p (x)} .

Commented by maxmathsup by imad last updated on 13/Feb/19

1) w e c a n u s e r e c u r r e n c e t o p r o v e t h a t f o r x \neq \bar{+} 1

P (x) = \frac{1 - x^{2^{n + 1}}}{1 - x^{2}}

2) P (z) = 0 \Leftrightarrow 1 - z^{2^{n + 1}} = 0 \Leftrightarrow 1 - z^{q} = 0 w i t h m = 2^{n + 1} t h e r o o t s o f

z^{m} = 1 a r e z_{k} = e^{\frac{i 2 k π}{m}} w i t h k \in [[0, m - 1]] \Rightarrow z_{k} = e^{\frac{i 2 k π}{2^{n + 1}}} = e^{\frac{i k π}{2^{n}}}

w i t h k \in [[0, 2^{n + 1} - 1]] b u t b u t e l i m i n a t e v a l u e s o f k / z_{k}^{2} = 1 .

Commented by maxmathsup by imad last updated on 13/Feb/19

P (x) = \prod_{k = 0_{z_{k} \neq \overset{- 1}{+}}}^{2^{n + 1} - 1} (x - z_{k})

Answered by mr W last updated on 12/Feb/19

1)

p (x) = (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})

(1 - x^{2}) p (x) = (1 - x^{2}) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})

(1 - x^{2}) p (x) = (1 - x^{4}) (1 + x^{4}) \dots (1 + x^{2^{n}})

(1 - x^{2}) p (x) = (1 - x^{8}) \dots (1 + x^{2^{n}})

\dots .

(1 - x^{2}) p (x) = (1 - x^{2^{n}}) \dots (1 + x^{2^{n}})

(1 - x^{2}) p (x) = (1 - x^{2^{n + 1}})

\Rightarrow p (x) = \frac{1 - x^{2^{n + 1}}}{1 - x^{2}}

Commented by maxmathsup by imad last updated on 12/Feb/19

c o r r e c t s i r b u t x \neq \bar{+} 1